# How would I go about proving this? [duplicate]

This question already has an answer here:

Question is:

Let $n$ represent a positive integer. Describe the largest set of values $n$ for which you think that $2^n < n!$

I'm not sure I get this question. For $n > 3$, it seems like $2^n$ is always less than $n!$ So how would I prove this description correct? Would I use principle of simple induction?

Thank you for your help.

## marked as duplicate by GNUSupporter 8964民主女神 地下教會, Hans Engler, B. Mehta, Wouter, mercioMay 11 '18 at 9:11

You can compare $2^n =2.2.2...$ , with $n!=1.2.3.4.....$ to show that once $n!$ is larger than $2^n$, it will always be larger than $2^n$, since $n!$ will be multiplied by numbers larger than $2$, while $2^n$ will grow by a precise factor of $2$.
A fairly simple induction argument will indeed allow you to prove that $2^n<n!$ for all $n>3$.
To show that the set of positive integers you obtained is the largest such set, you must show that $2^n\ge n!$ for all other positive integers $n.$